document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=82784>Question 82784</A>: <I><SPAN STYLE=\"word-break: break-all;\"> PROBLEM\r<BR>\n" );
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document.write( "A woman is 5 years older than her husband and 10 times as old as their daughter. In 14 years, the sum of their ages will be 100. How old is each one now?<BR>\n" );
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document.write( "Woman= h+5                   <BR>\n" );
document.write( "Husband= h                      <BR>\n" );
document.write( "Daughter=h+5/10(I think)         <BR>\n" );
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document.write( "          100 \r<BR>\n" );
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document.write( "(h+5)+(h)+(h+5/10)=100<BR>\n" );
document.write( "  2h+5+(h+5/10)=100<BR>\n" );
document.write( "    3h+10=1000<BR>\n" );
document.write( "    3h/3=990/3<BR>\n" );
document.write( "      h=330\r<BR>\n" );
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document.write( "My question is what did i do wrong I know that this is way too high of a number and I can't figure out where I went wrong.   Thanks<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #59386 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=59386\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> you have the age relationships right, but everyone will be 14 years older when their ages sum to 100\r<BR>\n" );
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document.write( "when you multiplied 2h+5+((h+5)/10)=100 by 10 you should have gotten 20h+50+h+5=1000 or 21h+55=1000\r<BR>\n" );
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document.write( "if you use the daughter's age as a base, then you avoid the fractions that led to your calculation error\r<BR>\n" );
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document.write( "(d+14)+(10d+14)+(10d-5+14)=100 ... 21d+37=100 ... d=3 so w=30 and h=25 </SPAN>\n" );
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